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Question Number 119292 by Algoritm last updated on 23/Oct/20
Answered by 1549442205PVT last updated on 24/Oct/20
Commented by 1549442205PVT last updated on 25/Oct/20
Denote by r the radius of two small circles CQ=x,PN=OM=2n;R and R_1 −the radius of semicircle and the great circle QO=QP=m⇒R_1 =R/2 Then we have relations: { ((x(R−x)=m^2 (1)(QC.QE=QP^2 ))),((((2r)/R)=(n/m)(2)(IN//OD))),(((x−r)^2 +n^2 =r^2 (3)(ΔOSF is right at S))),(((r/R)=((x−r)/x)(4)(((MF)/(MC))=((FS)/(QC))))) :} (3)⇔x^2 −2rx+n^2 =0(5) (1)⇒x^2 =Rx−m^2 .Replace into (5)we get Rx−m^2 −2rx+n^2 =0⇒x=((m^2 −n^2 )/(R−2r))(6) (2)⇒(R/(2m))=(r/n)⇒(R^2 /(4m^2 ))=(r^2 /n^2 )=((R^2 /4−r^2 )/(m^2 −n^2 ))(7) (4)⇒rx=Rx−Rr⇒x=((Rr)/(R−r))(8) Replace into (6)we get m^2 −n^2 =(((R−2r)Rr)/(R−r)) replace into (7) we get n^2 =((4(R−2r)Rr^3 )/((R−r)(R^2 −4r^2 )))(9) (8)⇒x−r=(r^2 /(R−r))(10).Replace (9)(10) into (3)we get (r^4 /((R−r)^2 ))+((4(R−2r)Rr^3 )/((R−r)(R^2 −4r^2 )))=r^2 ⇔r^2 (R^2 −4r^2 )+4(R−2r)(R−r)Rr =(R−r)^2 (R^2 −4r^2 ) ⇔R^2 r^2 −4r^4 +4R^3 r+8Rr^3 −12R^2 r^2 =R^4 −2R^3 r+R^2 r^2 −4R^2 r^2 +8Rr^3 −4r^4 ⇔R^4 −6R^3 r+8R^2 r^2 =0 8r^2 −6Rr+R^2 =0 Δ′=9R^2 −8R^2 =R^2 ⇒r=((3R−R)/8)=(R/4) (The root r=((3R+R)/8)=(R/2) is rejected) ⇒r=(R/4),replace into (8)we get x=(R/3) replace into (9)we get n=((R(√2))/6) ⇒m=((R(√2))/3)⇒cosMCQ^(�) =(x/R)=(1/3) ⇒MCN^(�) =2cos^(−1) ((1/3))⇒S_(sectorMCN) = ((πR^2 2cos^(−1) ((1/3)))/(2π))=R^2 cos^(−1) ((1/3)) S_(ΔMCN) =(m+2n)x=((2R^2 (√2))/9) ⇒S_(segMN) =R^2 cos^(−1) ((1/3))−((2R^2 (√2))/9) Since ΔNIP⋍ΔNCM with similar ratio equal to (1/4),S_(segMO) +S_(segNP) =2×(1/(16))S_(segMN) ΔODP⋍ΔMCN with similar ratio (1/2) ⇒S_(segOP) =(1/4)S_(segMN) ⇒S_(segOEP) =S_((D)) −S_(segOP) =((πR^2 )/4)−(1/4)S_(segMN) ⇒S_(green) =S_(segMN) −(1/8)S_(segMN) −(((πR^2 )/4)−(1/4)S_(segMN) ) =(R^2 /8)(9cos^(−1) ((1/3))−2(√2)−2π) ≈0.2458R^2
$$\mathrm{Denote}\:\mathrm{by}\:\mathrm{r}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{two}\:\mathrm{small}\:\mathrm{circles} \\ $$$$\mathrm{CQ}=\mathrm{x},\mathrm{PN}=\mathrm{OM}=\mathrm{2n};\mathrm{R}\:\mathrm{and}\:\mathrm{R}_{\mathrm{1}} −\mathrm{the} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{semicircle}\:\mathrm{and}\:\mathrm{the}\:\mathrm{great}\:\mathrm{circle} \\ $$$$\mathrm{QO}=\mathrm{QP}=\mathrm{m}\Rightarrow\mathrm{R}_{\mathrm{1}} =\mathrm{R}/\mathrm{2} \\ $$$$\mathrm{Then}\:\mathrm{we}\:\mathrm{have}\:\mathrm{relations}: \\ $$$$\begin{cases}{\mathrm{x}\left(\mathrm{R}−\mathrm{x}\right)=\mathrm{m}^{\mathrm{2}} \left(\mathrm{1}\right)\left(\mathrm{QC}.\mathrm{QE}=\mathrm{QP}^{\mathrm{2}} \right)}\\{\frac{\mathrm{2r}}{\mathrm{R}}=\frac{\mathrm{n}}{\mathrm{m}}\left(\mathrm{2}\right)\left(\mathrm{IN}//\mathrm{OD}\right)}\\{\left(\mathrm{x}−\mathrm{r}\right)^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \left(\mathrm{3}\right)\left(\Delta\mathrm{OSF}\:\mathrm{is}\:\mathrm{right}\:\mathrm{at}\:\mathrm{S}\right)}\\{\frac{\mathrm{r}}{\mathrm{R}}=\frac{\mathrm{x}−\mathrm{r}}{\mathrm{x}}\left(\mathrm{4}\right)\left(\frac{\mathrm{MF}}{\mathrm{MC}}=\frac{\mathrm{FS}}{\mathrm{QC}}\right)}\end{cases} \\ $$$$\left(\mathrm{3}\right)\Leftrightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{2rx}+\mathrm{n}^{\mathrm{2}} =\mathrm{0}\left(\mathrm{5}\right) \\ $$$$\left(\mathrm{1}\right)\Rightarrow\mathrm{x}^{\mathrm{2}} =\mathrm{Rx}−\mathrm{m}^{\mathrm{2}} .\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{5}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{Rx}−\mathrm{m}^{\mathrm{2}} −\mathrm{2rx}+\mathrm{n}^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{x}=\frac{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }{\mathrm{R}−\mathrm{2r}}\left(\mathrm{6}\right) \\ $$$$\left(\mathrm{2}\right)\Rightarrow\frac{\mathrm{R}}{\mathrm{2m}}=\frac{\mathrm{r}}{\mathrm{n}}\Rightarrow\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{4m}^{\mathrm{2}} }=\frac{\mathrm{r}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }=\frac{\mathrm{R}^{\mathrm{2}} /\mathrm{4}−\mathrm{r}^{\mathrm{2}} }{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} }\left(\mathrm{7}\right) \\ $$$$\left(\mathrm{4}\right)\Rightarrow\mathrm{rx}=\mathrm{Rx}−\mathrm{Rr}\Rightarrow\mathrm{x}=\frac{\mathrm{Rr}}{\mathrm{R}−\mathrm{r}}\left(\mathrm{8}\right) \\ $$$$\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{6}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} =\frac{\left(\mathrm{R}−\mathrm{2r}\right)\mathrm{Rr}}{\mathrm{R}−\mathrm{r}}\:\mathrm{replace}\:\mathrm{into}\:\left(\mathrm{7}\right)\:\mathrm{we}\: \\ $$$$\mathrm{get}\:\mathrm{n}^{\mathrm{2}} =\frac{\mathrm{4}\left(\mathrm{R}−\mathrm{2r}\right)\mathrm{Rr}^{\mathrm{3}} }{\left(\mathrm{R}−\mathrm{r}\right)\left(\mathrm{R}^{\mathrm{2}} −\mathrm{4r}^{\mathrm{2}} \right)}\left(\mathrm{9}\right) \\ $$$$\left(\mathrm{8}\right)\Rightarrow\mathrm{x}−\mathrm{r}=\frac{\mathrm{r}^{\mathrm{2}} }{\mathrm{R}−\mathrm{r}}\left(\mathrm{10}\right).\mathrm{Replace}\:\left(\mathrm{9}\right)\left(\mathrm{10}\right) \\ $$$$\mathrm{into}\:\left(\mathrm{3}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{r}^{\mathrm{4}} }{\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} }+\frac{\mathrm{4}\left(\mathrm{R}−\mathrm{2r}\right)\mathrm{Rr}^{\mathrm{3}} }{\left(\mathrm{R}−\mathrm{r}\right)\left(\mathrm{R}^{\mathrm{2}} −\mathrm{4r}^{\mathrm{2}} \right)}=\mathrm{r}^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{r}^{\mathrm{2}} \left(\mathrm{R}^{\mathrm{2}} −\mathrm{4r}^{\mathrm{2}} \right)+\mathrm{4}\left(\mathrm{R}−\mathrm{2r}\right)\left(\mathrm{R}−\mathrm{r}\right)\mathrm{Rr} \\ $$$$=\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} \left(\mathrm{R}^{\mathrm{2}} −\mathrm{4r}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow\mathrm{R}^{\mathrm{2}} \mathrm{r}^{\mathrm{2}} −\mathrm{4r}^{\mathrm{4}} +\mathrm{4R}^{\mathrm{3}} \mathrm{r}+\mathrm{8Rr}^{\mathrm{3}} −\mathrm{12R}^{\mathrm{2}} \mathrm{r}^{\mathrm{2}} \\ $$$$=\mathrm{R}^{\mathrm{4}} −\mathrm{2R}^{\mathrm{3}} \mathrm{r}+\mathrm{R}^{\mathrm{2}} \mathrm{r}^{\mathrm{2}} −\mathrm{4R}^{\mathrm{2}} \mathrm{r}^{\mathrm{2}} +\mathrm{8Rr}^{\mathrm{3}} −\mathrm{4r}^{\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{R}^{\mathrm{4}} −\mathrm{6R}^{\mathrm{3}} \mathrm{r}+\mathrm{8R}^{\mathrm{2}} \mathrm{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{8r}^{\mathrm{2}} −\mathrm{6Rr}+\mathrm{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta'=\mathrm{9R}^{\mathrm{2}} −\mathrm{8R}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} \Rightarrow\mathrm{r}=\frac{\mathrm{3R}−\mathrm{R}}{\mathrm{8}}=\frac{\mathrm{R}}{\mathrm{4}} \\ $$$$\left(\mathrm{The}\:\mathrm{root}\:\mathrm{r}=\frac{\mathrm{3R}+\mathrm{R}}{\mathrm{8}}=\frac{\mathrm{R}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{rejected}\right) \\ $$$$\Rightarrow\mathrm{r}=\frac{\mathrm{R}}{\mathrm{4}},\mathrm{replace}\:\mathrm{into}\:\left(\mathrm{8}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{x}=\frac{\mathrm{R}}{\mathrm{3}} \\ $$$$\mathrm{replace}\:\mathrm{into}\:\left(\mathrm{9}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{n}=\frac{\mathrm{R}\sqrt{\mathrm{2}}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{m}=\frac{\mathrm{R}\sqrt{\mathrm{2}}}{\mathrm{3}}\Rightarrow\mathrm{cos}\widehat {\mathrm{MCQ}}=\frac{\mathrm{x}}{\mathrm{R}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\widehat {\mathrm{MCN}}=\mathrm{2cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Rightarrow\mathrm{S}_{\mathrm{sectorMCN}} = \\ $$$$\frac{\pi\mathrm{R}^{\mathrm{2}} \mathrm{2cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{2}\pi}=\mathrm{R}^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\mathrm{S}_{\Delta\mathrm{MCN}} =\left(\mathrm{m}+\mathrm{2n}\right)\mathrm{x}=\frac{\mathrm{2R}^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{segMN}} =\mathrm{R}^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{2R}^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{9}} \\ $$$$\mathrm{Since}\:\Delta\mathrm{NIP}\backsimeq\Delta\mathrm{NCM}\:\mathrm{with}\:\mathrm{similar}\:\mathrm{ratio}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\frac{\mathrm{1}}{\mathrm{4}},\mathrm{S}_{\mathrm{segMO}} +\mathrm{S}_{\mathrm{segNP}} =\mathrm{2}×\frac{\mathrm{1}}{\mathrm{16}}\mathrm{S}_{\mathrm{segMN}} \\ $$$$\Delta\mathrm{ODP}\backsimeq\Delta\mathrm{MCN}\:\mathrm{with}\:\mathrm{similar}\:\mathrm{ratio}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{segOP}} =\frac{\mathrm{1}}{\mathrm{4}}\mathrm{S}_{\mathrm{segMN}} \Rightarrow\mathrm{S}_{\mathrm{segOEP}} =\mathrm{S}_{\left(\mathrm{D}\right)} −\mathrm{S}_{\mathrm{segOP}} \\ $$$$=\frac{\pi\mathrm{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{S}_{\mathrm{segMN}} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{green}} =\mathrm{S}_{\mathrm{segMN}} −\frac{\mathrm{1}}{\mathrm{8}}\mathrm{S}_{\mathrm{segMN}} −\left(\frac{\pi\mathrm{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{S}_{\mathrm{segMN}} \right) \\ $$$$=\frac{\mathrm{R}^{\mathrm{2}} }{\mathrm{8}}\left(\mathrm{9}\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\pi\right)\:\approx\mathrm{0}.\mathrm{2458R}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 24/Oct/20
Commented by mr W last updated on 24/Oct/20
OP=PC=(R/2) OQ=R−r PQ=(R/2)+r OE^2 =OQ^2 −QE^2 =(R−r)^2 −r^2 =R(R−2r) =PQ^2 −(OP−QE)^2 =((R/2)+r)^2 −((R/2)−r)^2 =2Rr ⇒2Rr=R(R−2r) ⇒r=(R/4) sin α=(r/(R−r))=(1/3) cos γ=(((R/2)−r)/((R/2)+r))=(1/3)=cos φ ⇒φ=γ=(π/2)−α 2β=π−(φ+γ) ⇒β=(π/2)−φ=α (A_(green) /2)=((R^2 φ)/2)−2×((r^2 φ)/2)−((((R/2))^2 (π−γ))/2)−(((R/2)×(√(R(R−2r))))/2) =((R^2 (9φ−2π))/(16))−(((√2)R^2 )/8) =((R^2 (9 cos^(−1) (1/3)−2π−2(√2)))/(16)) A_(green) =((R^2 (9 cos^(−1) (1/3)−2π−2(√2)))/8)≈0.2459R^2
$${OP}={PC}=\frac{{R}}{\mathrm{2}} \\ $$$${OQ}={R}−{r} \\ $$$${PQ}=\frac{{R}}{\mathrm{2}}+{r} \\ $$$${OE}^{\mathrm{2}} ={OQ}^{\mathrm{2}} −{QE}^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} ={R}\left({R}−\mathrm{2}{r}\right) \\ $$$$={PQ}^{\mathrm{2}} −\left({OP}−{QE}\right)^{\mathrm{2}} =\left(\frac{{R}}{\mathrm{2}}+{r}\right)^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{2}}−{r}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}{Rr} \\ $$$$\Rightarrow\mathrm{2}{Rr}={R}\left({R}−\mathrm{2}{r}\right) \\ $$$$\Rightarrow{r}=\frac{{R}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\alpha=\frac{{r}}{{R}−{r}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\gamma=\frac{\frac{{R}}{\mathrm{2}}−{r}}{\frac{{R}}{\mathrm{2}}+{r}}=\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{cos}\:\phi\:\Rightarrow\phi=\gamma=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$$\mathrm{2}\beta=\pi−\left(\phi+\gamma\right)\: \\ $$$$\Rightarrow\beta=\frac{\pi}{\mathrm{2}}−\phi=\alpha \\ $$$$\frac{{A}_{{green}} }{\mathrm{2}}=\frac{{R}^{\mathrm{2}} \phi}{\mathrm{2}}−\mathrm{2}×\frac{{r}^{\mathrm{2}} \phi}{\mathrm{2}}−\frac{\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\pi−\gamma\right)}{\mathrm{2}}−\frac{\frac{{R}}{\mathrm{2}}×\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}}{\mathrm{2}} \\ $$$$=\frac{{R}^{\mathrm{2}} \left(\mathrm{9}\phi−\mathrm{2}\pi\right)}{\mathrm{16}}−\frac{\sqrt{\mathrm{2}}{R}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$=\frac{{R}^{\mathrm{2}} \left(\mathrm{9}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}−\mathrm{2}\pi−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{16}} \\ $$$${A}_{{green}} =\frac{{R}^{\mathrm{2}} \left(\mathrm{9}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}−\mathrm{2}\pi−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{8}}\approx\mathrm{0}.\mathrm{2459}{R}^{\mathrm{2}} \\ $$

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